Đáp án:
\(\begin{array}{l}
b)\\
{m_{Fe}} = 14g\\
c)\\
C{\% _{{H_2}S{O_4}}} = 4,9\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,25mol\\
{m_{Fe}} = n \times M = 0,25 \times 56 = 14g\\
c)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,25mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,25 \times 98 = 24,5g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{24,5}}{{500}} \times 100\% = 4,9\%
\end{array}\)